I've been trying and trying... but there does not seem to be three

Number Theory Level 5

Consider the expression $\sqrt {\frac {(n+1)(2n+1)}{6}}$ .

For some whole numbers $n$ , the expression is also a whole number. Let the three smallest such whole numbers be $a_{1}, a_{2}, a_{3}$ . Find the value of $\sum_{i=1}^3 a_{i}$ .

The answer is 65859.

2 solutions

John Ashley Capellan
Sep 29, 2014

Express the numerator of the radicand as (n + 1)(2n + 1) = 2n^2 + 3n + 1 = 6k^2. Then express it in the form: 2(n^2 + 3n/2 + 1/2) = 6k^2 and n^2 + 3n/2 + 1/2 = 3k^2. Performing the completing the square, this gives (n + 3/4)^2 - 1/16 = 3k^2 or (4n + 3)^2 / 16 - 1/16 = 3k^2 or (4n + 3)^2 - 1 = 48k^2. Here, we solve a form of Pell's equation with (4n + 3)^2 - 48k^2 = 1. Using the substitution a = 4n + 3 and b = k. We solve the Pell equation: a^2 - 48k^2 = 1. Using Brute force for the form (7 + 4 sqrt(3))^n, we get the three smallest solutions (after checking..) 1, 337, 65521.

Saurabh Kataria
Sep 19, 2014

The numbers are 1, 337 and 65521. The first number is easy to check manually. However, for the next two you need to write a computer program which evaluates the expression for every whole number successively. You can use an infinite loop and in C++, you can use std::modf function to check whether the double number (that has been evaluated using the expression) is an integer or not. I really believe there must be a non-brute force approach to this problem. I will be happy to see if someone can provide that.

this is a NUMBER THEORY question.Not CS!!

Adarsh Kumar - 6 years, 8 months ago

Please re-read the last line in my solution.

Saurabh Kataria - 6 years, 8 months ago

Yeah. What about it?

Adarsh Kumar - 6 years, 8 months ago

@Adarsh Kumar – You are level 5 in number theory, why aren't you writing a solution then ?

Saurabh Kataria - 6 years, 8 months ago

I've been trying and trying... but there does not seem to be three

The answer is 65859.

2 solutions

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